What is the volume of 4.5 g of "CO_2"?

2 Answers
Dec 1, 2015

The volume of "4.5 g CO"_2" is "2300 cm"^3" at STP.

Explanation:

At STP, "273.15 K" and "1 atm", the density of carbon dioxide is "0.001977 g/cm"^3".
http://www.engineeringtoolbox.com/gas-density-d_158.html

We can use the given mass and known density to determine the volume of "CO"_2" at STP. Divide the mass by the density.

4.5cancel"g CO"_2xx(1"cm"^3 "CO"_2)/(0.001977cancel"g CO"_2)="2300 cm"^3 "CO"_2" (rounded to two significant figures)

Dec 1, 2015

The volume of "4.5 g CO"_2" is "2300 cm"^3" at STP.

Explanation:

We can also use the ideal gas law to solve this problem.
The equation is PV=nRT, where n is the number of moles, and R is the gas constant.

In this problem, T and P are at STP, "273.15 K" and "1 atm".

Determine moles of "CO"_2" by dividing the given mass by the molar mass.

4.5cancel"g CO"_2xx(1"mol CO"_2)/(44.01cancel"g CO"_2)="0.10225 mol CO"_2"

I am leaving some guard digits to reduce rounding errors.

Ideal Gas Law

Given/Known
P="1 atm"
n="0.10225 mol"
R="0.082057338 L atm K"^(-1) "mol"^(-1)"
https://en.m.wikipedia.org/wiki/Gas_constant
T="273.15 K"

Unknown
Volume, V

Equation
P_1V_1=nRT

Solution
Rearrange the equation to isolate V and solve.

V=(nRT)/(P)

V=((0.10225cancel"mol"xx0.082057338 "L" cancel"atm" cancel("K"^(-1)) cancel("mol"^(-1))xx273.15cancel"K"))/(1cancel"atm")="2.3 L" (rounded to two significant figures)

Convert to "cm"^3".

2.3cancel"L"xx(1000cancel"mL")/(1cancel"L")xx(1"cm"^3)/(1cancel"mL")="2300 cm"^3"